A 27.6 ml volume of chlorine gas is collected over water at 23 C and 750 torr. The mass of the gas is 78.4 mg. The vapor pressure of water is 21 torr.

a) What is the pressure of the chlorine gas?

b) How volume does the chlorine gas occupy?

c) How many moles of chlorine gas are collected?

a). Total P = partial pressure chloring + partial pressure water vapor.

You know total P and you know parital pressure of water vapor. Solve for partial pressure chlorine.
b)I may have missed something but don't they tell you the volume is 27.6 mL?
c)Use PV = nRT
Post your work if you need additional assistance.

To calculate the pressure of the chlorine gas, we need to subtract the vapor pressure of water from the total pressure.

a) Pressure of chlorine gas = Total pressure - Vapor pressure of water
Pressure of chlorine gas = 750 torr - 21 torr
Pressure of chlorine gas = 729 torr

b) The volume of the chlorine gas can be calculated using the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(K·mol)), and T is the temperature in Kelvin.

First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(C) + 273.15
T(K) = 23°C + 273.15
T(K) = 296.15 K

Now we can rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P

Substituting the known values:
V = (n * 0.0821 L·atm/(K·mol) * 296.15 K) / 729 torr

The pressure and temperature units need to be adjusted to match the units in the ideal gas constant:
V = (n * 0.0821 L·atm/(K·mol) * 296.15 K) / (729 torr * (1 atm / 760 torr))

Simplifying the equation:
V = (n * R * T) / P

Since we are solving for volume, we can rearrange the equation:
V = (78.4 mg / (35.45 g/mol)) * (0.0821 L·atm/(K·mol) * 296.15 K) / 729 torr

The mass needs to be converted to moles using the molar mass of chlorine gas, which is approximately 35.45 g/mol.
V = (2.212 * 10^-3 mol) * (0.0821 L·atm/(K·mol) * 296.15 K) / 729 torr

Finally, we can plug in the values and calculate the volume:
V = (0.648042 L) / (729 torr)

V ≈ 0.0009 L
V ≈ 0.9 mL

Therefore, the volume of the chlorine gas is approximately 0.9 mL.

c) To calculate the number of moles of chlorine gas, we can use the equation:
n = mass / molar mass

n = 78.4 mg / 35.45 g/mol
n = 2.212 * 10^-3 mol

Therefore, approximately 2.212 * 10^-3 moles of chlorine gas are collected.

To answer these questions, we need to use the ideal gas law equation, which relates the pressure, volume, temperature, and number of moles of a gas.

The ideal gas law equation is: PV = nRT

Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin (K)

Let's solve each question step by step.

a) What is the pressure of the chlorine gas?
To find the pressure of the chlorine gas, we need to subtract the vapor pressure of water from the total pressure. The total pressure is given as 750 torr, and the vapor pressure of water is given as 21 torr. Therefore, the pressure of the chlorine gas is:
Chlorine gas pressure = Total pressure - Vapor pressure of water
= 750 torr - 21 torr
= 729 torr

b) How volume does the chlorine gas occupy?
The volume of the chlorine gas can be calculated using the ideal gas law equation. Rearranging the equation, we get:
V = (nRT) / P

Given:
n (number of moles of chlorine gas) = ?
R (ideal gas constant) = 0.0821 L·atm/(mol·K)
P (pressure of chlorine gas) = 729 torr
T (temperature in Kelvin) = 23°C + 273.15 = 296.15 K

We need to find the number of moles of chlorine gas (n). Rearranging the ideal gas law equation again, we get:
n = (PV) / (RT)

Substituting the known values into the equation:
n = (729 torr * 27.6 ml) / (0.0821 L·atm/(mol·K) * 296.15 K)

Converting ml to L by dividing by 1000 (1 L = 1000 ml):
n = (729 torr * 0.0276 L) / (0.0821 L·atm/(mol·K) * 296.15 K)
n = 0.0196 mol

Now, we can substitute the value of n (0.0196 mol) back into the volume equation to find the volume of the chlorine gas:
V = (nRT) / P
V = (0.0196 mol * 0.0821 L·atm/(mol·K) * 296.15 K) / 729 torr
V = 0.023 L or 23.0 ml (rounded to two decimal places)

c) How many moles of chlorine gas are collected?
We have already calculated the number of moles of chlorine gas in the previous step, which is 0.0196 mol.

So, the answer is:
a) The pressure of the chlorine gas is 729 torr.
b) The volume of the chlorine gas is 23.0 ml.
c) The number of moles of the chlorine gas collected is 0.0196 mol.